# An equilateral triangle is inscribed in the parabola y2 = 4 ax,

Question:

An equilateral triangle is inscribed in the parabola $y^{2}=4 \mathrm{ax}$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution:

Let $O A B$ be the equilateral triangle inscribed in parabola $y^{2}=4 a x$.

Let $\mathrm{AB}$ intersect the $x$-axis at point $\mathrm{C}$.

Let $\mathrm{OC}=k$

From the equation of the given parabola, we have $y^{2}=4 a k \Rightarrow y=\pm 2 \sqrt{a k}$

$\therefore$ The respective coordinates of points A and B are $(k, 2 \sqrt{a k})$, and $(k,-2 \sqrt{a k})$

$\mathrm{AB}=\mathrm{CA}+\mathrm{CB}=2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}$

Since $O A B$ is an equilateral triangle, $O A^{2}=A B^{2}$.

$\therefore k^{2}+(2 \sqrt{a k})^{2}=(4 \sqrt{a k})^{2}$

$\Rightarrow k^{2}+4 a k=16 a k$

$\Rightarrow k^{2}=12 a k$

$\Rightarrow k=12 a$

$\therefore \mathrm{AB}=4 \sqrt{a k}=4 \sqrt{a \times 12 a}=4 \sqrt{12 a^{2}}=8 \sqrt{3} a$

Thus, the side of the equilateral triangle inscribed in parabola $y^{2}=4 a x$ is $8 \sqrt{3} a$.